Equation Of First Order
And
First Degree
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Separation of variables
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If in an equation, it is possible to get all the functions of x and dx in to one side and all the function of y and dy to the other, the variable are said to be separable.
Working rule to solve an equation in which variables are separable.
step 1 )
Let dy/dx =f₁(x) f₂(y) ..(1)
be given equation f₁(x) is a function of x alone and f₂(y) is a function of y alone.
step 2)
From (1) separating variables,
[1/f₂(y)] dy = f₁(x) dx ............(2)
step 3)
Integrating both sides of (2), we have ∫[1/f₂(y)] dy=∫f₁(x)dx + c...(3)
where c is constant of integration, is the required solution.
Note 1:
In all solution (3), an arbitrary constant c must be added in any one side only. If c is not added, then the solution obtained will not be a general solution of (1).
Note 2: To simplify the solution (3), the constant of integration can be chosen in any suitable form so as to get the final solution in a form as simple as possible. Accordingly, we are write log c, tan⁻¹ c, sin c, eᶜ,
(1/2). C , (-1/3). C etc in place of c in some solutions.
Note 3 :
The students are advised to remember by heart the following formulas. These will help them to write solution (3) in compact form
i) log x+ log y= log xy
ii) log x - log y = log(x/y)
iii) n log x = log xⁿ
iv)tan⁻¹x+tan⁻¹y= tan⁻¹[(x+y)/(1-xy)]
v) tan⁻¹x-tan⁻¹y = tan⁻¹[(x-y)/(1+xy)]
vi) eˡᵒᵍ ᶠ⁽ˣ⁾ = f(x).
EXAMPLE
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1) dy/dx = eˣ⁻ʸ + x² e⁻ʸ
dy/dx = e⁻ʸ( eˣ +x²)
or eʸ dy = (x² +eˣ) dx
integrating eʸ = x³/3 + eˣ + c, c is con.
2) √(1+x²+y²+x²y²) + xy (dy/dx) = 0
=√{(1+x²)(1+y²)} + xy(dy/dx) = 0
= √(1+x²)dx /x + ydy/√(1+y²) =0
=(1+x²)dx/x√(1+x²)+ydy/√(1+y²)
=0
= ∫ dx/x√(1+x²) +∫xdx/√(1+x²)+
ydy/(1+y²) =c
= log x - log{1- √(1+x²)}+
√(1+x²)+√(1+y²) =c
EXERCISE
*********
1) dy/dx = eˣ⁺ʸ + x² eʸ
2) (dy/dx)tan y=sin(x + y) + sin(x-y)
3) dy/dx=(sinx+xcosx)/
{y(2log y+1)}
4) dy/dx={x(2log x+1)}/
(siny +ycosy)
5) log(dy/dx) = ax + by
6) y - x(dy/dx) = a(y² + dy/dx)
7) 3eˣ tan y dx + (1- eˣ)sec² y dy =0
8) ₑx+y dy ₌ ₓ² ₑx³+y dx
9) dy/dx = eˣ ⁺ ʸ when x=1, y=1. find y when x= -1
10) (eˣ + 1)y0 dy = (y +1)eˣ dx
11) (dy/dx) - y tan x = - y sec² x
12) x√(1+y²) dx + y√(1+x²) dy =0
13) (2ax+x²)(dy/dx) = a² + 2ax
14) dr = a (r sinθ dθ - cosθ dr)
15) (eʸ +1) cosx dx + eʸsin x dy = 0
16) √(a+x) (dy/dx) +x = 0
17) dy/dx = √{(1-y²)/(1 -x²)}
18) (x²-yx²)dy + (y²+xy²) dx =0
19) (xy² +x)dx + (yx² +y) dy = 0
20) sec²x tany dx+sec²y tan xdy =0
21) (1+x)y dx + (1+y)x dx = 0
22) (1- x²)(1 - y) dx = xy(1+y)dx
23) x²(y+1)dx + y²(x - 1) dy =0
24) (dy/dx)tan y=sin(x+y)+sin(x - y)
25) y - x dy/dx = 3(1+ x² dy/dx)
26) cosy log(secx +tanx) dx =
cos xlog(sec y + tan y) dy
27) x dy - y dx = (a² + y²)¹/² dx
MISCELLANEOUS PROBLEM
1) Find the curves passing through (0,1) and satisfying sin(dy/dx)
2) Find the function f which satisfies the equation df/dx = 2f, given that f(0) = e³
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VARIABLE UNSEPARABLE
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Transformation of some equations in the form in which variables are separable Equations of the form
dy/dx = f(ax+ by+ c). OR
dy/dx = f(ax + by)
can be reduced to an Equation in which variables can be separated.
For this purpose, we use the substitution ax +by+c = v OR
ax + by=v
EXAMPLE (1)
dy/dx = (4x + y +1)²
Let 4x + y +1 = v. ........(1)
Differentiating (1) with respect to x, we get 4 + (dy/dx)= dv/dx
OR
dy/dx=(dv/dx) - 4 ........(2)
Using (1) and (2),
the equation becomes
(dv/dx) - 4 = v² OR
dv/dx = 4 + v²
Now , separating variables x and v,
So dx =dv/(4+v²)
Integrating,
x + c ' = (1/2).Tan⁻¹(v/2),
where c ' is an arbitrary constant.
Or, 2x + c = Tan⁻¹(v/2)
Or, v= 2 tan(2x +c), Where c = 2c '
Or, 4x + y + 1=2 tan(2x +c),
using (1)
EXAMPLE (2)
(x+y)² (dy/dx) = a².
Let x+y=v .......(1)
Differentiating 1+(dy/dx) OR
dy/dx= dv/dx - 1 .......(2)
Using (1) and (2) the given Equation becomes
v²(dv/dx -1)= a² OR
v²= dv/dx = a²+v² OR
dx=v²/(v²+a²) OR
dx = [1- a²/(a²+v²)]
Integrating,
x+c = v - a² (1/a) Tan⁻¹(v/a),
where c is arbitrary constant.
Or x +c= x+y - Tan⁻¹{(x+y)/2}
Or y - a Tan⁻¹{(x+y)/a}= c
EXERCISE
***********
1) dy/dx=sec(x+y)
or
cos(x+y)dy=dx
2) dy/dx= sin(x+y) + cos(x +y)
3) (x+y)(dx - dy) = dx+ dy
4) dy/dx = (4x +6y +5)/(3y +2x +4)
5) (x+2y -1)dx = (x + 2y +1) dy
6) dy/dx= (x +y)²
7) dy/dx +1 = eˣ⁺ʸ
8) (2x +y +1) dx + (4x +2y -1) dy= 0
9) (x - y - 2)dx - (2x - 2y -3) dy =0
10) (x +y +1) (dy/dx) = 1
11) sin⁻¹(dy/dx) = x +y
12) (2x + 4y +3)(dy/dx) = 2y +x + 1
13) (4x+6y+5)/(3y+2x+4)(dy/dx)=1
14) dy/dx= (x-y+3)/(2x - 2y +5)
15) (2x +2y +3)dy - (x+y+1) dx =0
16) (x -y)² (dy/dx) = a²
17) (x+y-a)/(x+y-b)(dy/dx) =
(x+y+a)/(x+y+b)
18) dy/dx = cos (x+y)
19) dy/dx= eˣ⁺ʸ given x=1, y=1,
prove y(-1)= -1
20) dy/dx= (x+y+1)/(x+y-1)
when y= 1/3, at x= 2/3
21) (x+y-1)dy = (x+y) dx
22) dy/dx = (x-y+3)/(2x-2y+5)
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HOMOGENEOUS EQUATION
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Definition)
A differential equation of first order and first degree is said to be homogeneous if it can be put in the form dy/dx = f(y/x)
Working Rule)
Let the given equation be homogeneous. then , by definitiin, the given equation can be put in the form dy/dx =f(y/x) .......(1)
To solve(1),let y/x= v i.e., y=vx ..(2)
Differentiating w.r.t.x, (2)
dy/dx= v+x(dv/dx ...(3)
Using (2) and (3), (1) becomes
v+ x dv/dx = f(v) or x dv/dx=f(v) -v
separating the variables x and v, we have dx/x =ln(dv/{f(v) -v}
so that
log x + c = dv/{f(v) -v} where c is an arbitrary constant. after integratiin, replace v by y/x.
Examples
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1) (x³+3xy²)dx+(y³+3x²y)dy =0
given dy/dx = -(x³+3xy²)/(y³+3x²y)
dy/dx ={1+3(y/x)²}/{(y/x)³+3(y/x)}
take y/x =v, i.e., y=vx
so that dy/dx = v+ x (dv/dx)
so v+ x dv/dx = -(1+3v²)/(v³+3v)
or x dv/dx = -(1+3v²)/(v³ +3v) - v
= -(v⁴ +6v² +1)/(v³ +3v)
or 4dx/x=-(4v³ +12v)/
(v⁴ + 6v² +1)dv
integrating 4 log x= - log(v⁴+6v²+1) + log c, c being arbitrary constant.
or log x⁴ = log[c/(v⁴+6v² +1)], i.e.,
x⁴(v⁴+6v²+1)=c
or y⁴ +6x²y²+x⁴ +c
or (x²+y²)² +4x²y² =c as y/x =v
EXERCISE
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1) (x² +y²)dx - 2xydy =0
2) y² +x² (dy/dx) = xy(dy/dx)
3) (x² +xy)dy = (x²+ y²) dx
4) dy/dx = y/x + sin(y/x)
5) (x² +y²) (dy/dx) = xy
6) (x² -y²) dy = 2xy dx
7) (x³ - y³)dx + xy² dy =0
8) y² dx + (xy +x²) dy =0
9) x(dy/dx) + (y²/x)= y
10) x²y dx - (x³+ y³) dy = 0
11) (x +y) dy + (x - y ) dx = 0
or
y - x((dy/dx) = x + y(dy/dx)
12) x(x - y)dy + y² dx =0
13) x(x - y) dy = y(x + y)dx
14) xsin (y/x) (dy/dx)= ysin (y/x) - x
15) x² dy + y(x+ y)dx =0
16) (x³ - 3xy²)dx = (y³ - 3x²y) dy
17) 2 (dy/dx) = [y(x + y)/x²]
or
2(dy/dx) - (y/x) = y²/x²
18) (x³ - 2y³) dx + 3xy² dy =0
19) dy/dx = (xy² - x²y)/x³
20) (x² + y²) dx +2xy dy = 0
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EQUATION REDUCIBLE TO
HOMOGENEOUS FORM.
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Equ. dy/dx=(ax+by+c)/(a′x+by +c)
where a/a′ ≠ b/b′ .........(1)
can be reduced to homogenous as
Take x= X + h and y =Y+k ...(2)
where X and Y are new variables and h and k are constants to be chosen that the resulting Equation in terms of X and Y may become homogeneous.
From (2), dx=d X and dy= dY
so that dy/DX = dY/dX. ...(3)
Using (2) and (3), (1) becomes
dY/dX= {a(X+h)+b(Y+k)+c}
{a′(X +h)b′(Y+k)+c′}
= {aX +bY +(ah+bk+c)}
{a′X +b′Y +(a′h+b'k+c)} .....(4)
In order to make (4) homogeneous, chose h and k so as to satisfy the following two Equation ah + bk+c=0 and a'h +b'k+c' =0. ........(5)
Solving (5), h= (bc' -b'c)/(ab' - a'b) and k= (ca' - c'a)/(ab' - a'b) .......(6)
Given that a/a' ≠ b/b'. Therefore,
(ab' - a'b) ≠ 0. Hence, h and k given by (6) are meaningful, i.e., h and k will exist. Now, h and k are shown. So from (2), we get
X= x - h. And Y= y - k. .....(7)
In view of (5), (4) reduces to
dY/dX = (aX +bY)/(a'X + b'Y)
= {a+b(Y/X)}/{a'+b'(Y/X)
Which is surely homogeneous Equation in X and Y and can be solved by putting Y/X =v as usual. After getting solution in terms of X and Y, we remove X and Y by using (7) and obtain solution in terms of the original variables x and y.
EXAMPLE
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dY/dx = (x+2y -3)/(2x +y -3)
Let x= X+h, and y=Y+k
So dy/dx= dY/dX. ................(1)
So Equation becomes
dY/dX={X+2Y+(h+2k-3)}/
{2X +Y+(2h+k - 3)} ........(2)
Choose h,k so that
h+2k-3=0, and 2h+k - 3=0 ......(3)
Solving (3) we get h=1, k=1 so in(1) we have X=x - 1, and Y = y - 1.....(4)
Using (3) in (2), we get
dY/dX= (X+2Y)/(2X+Y)
= {1+(2Y/X)}/{2+(Y/X)} ...(5)
Take Y/X=v, i.e., Y=vx
dY/dX= v+X(d v/dX)...(6)
From (5) and (6), we have
v+X d v/dX=(1+2v)/(2+v)
Or X dv/dX = (1+2v)/(2+v) - v
= (1- v²)/(2+v)
Or dX/X ={(2+v)dv}/{(1 -v)(1+v)}
=[1/2 {1/(1+v)} +3/2{1/(1-v)}] dv,
Integrating
logX+logc=(1/2)[log(1+v)- 3log(1-v)]
Or 2 log (cX)= log(1+v)/(1-v)³
Or X²c² = (1+v)/(1 - v)³
Or X²c²(1-Y/X)³=1+ Y/X, as v= Y/X
Or c²(X-Y)³= X+Y
or c²{x-1-(y-1)}²= x-1+y-1,. ...by (4)
Or c'(x-y)²= x+y-2, taking c'= c².c' being an arbitrary constant.
EXERCISE
*************
1) dy/dx + (x-y-2)/(x-2y-3)=0
2) dy/dx= (x+y+4)/(x-y-6)
3) dy/dx= (x-2y+5)/(2x+y-1)
4) dy/dx= (x+y-2)/(y-x-4)
5) (2x²+3y²-7)x dx-(3x²+2y²-8)y dy=0
6) dy/dx= (x+2y+3)/(2x+3y+4)
7) dy/dx= (y-x-1)/(y+x+5)
8) dy/dx=(2x +2y -2)/(3x+y-5)
9) dy/dx= (2x-y+1)/(x+2y-3)
10) (x+2y-2)dx +(2x-y+3) dy=0
11) (2x+3y-5)(dy/dx +(3x+2y-5) =0
12) (x -y)dy = (x+y+1)dx
13)(6x+2y-10)(dy/dx)-2x-9y +20 =0
14) (6x -2y -7) dx = (2x +3y -6) dy
15) (3y -7x +7)dx+ (7y -3x +3) dy=0
16) (x-y-1)dx + (4y+x-1)dy =0
17) (2x +3y +4) dy = (x+2y +3) dx.
